Question 1

Two cars start at the same time from places A and B which are 60 km apart and move in the same direction. The car from A travels at 35 km/h and the car from B travels at 25 km/h. After how many hours will the faster car catch up with the slower car?

Accepted Answer

Introduction / Context: This question involves relative speed when two vehicles move in the same direction. Catch up problems are common in aptitude tests and help learners understand how the difference in speeds affects the time required for one object to overtake another. Here we are told that two cars start from different points 60 km apart but move in the same direction with different speeds. We must find the time it takes for the faster car to catch the slower one. Given Data / Assumptions: Distance between A and B = 60 km. Speed of the car starting from A = 35 km/h. Speed of the car starting from B = 25 km/h. Both cars start at the same time and move in the same direction. We assume straight line motion with constant speeds and no stops. Concept / Approach: When two objects move in the same direction, the effective or relative speed between them is the difference of their speeds. The faster car reduces the gap between them at this relative speed. Time required to catch up is equal to the initial distance between them divided by the relative speed. We identify which car is faster, compute the relative speed, and then use the formula time = distance / speed. Step-by-Step Solution: The faster car has speed 35 km/h and the slower car has speed 25 km/h. The relative speed when moving in the same direction = 35 - 25 = 10 km/h. Initial distance between the cars = 60 km. Time to catch up = distance / relative speed = 60 / 10 hours. Compute the time: 60 / 10 = 6 hours. Verification / Alternative check: We can verify by calculating how far each car has travelled after 6 hours. In 6 hours the faster car covers 35 * 6 = 210 km. In the same time the slower car covers 25 * 6 = 150 km. The difference between these distances is 210 - 150 = 60 km, which is exactly the initial separation. Therefore after 6 hours the faster car catches up to the slower car, confirming that 6 hours is correct. Why Other Options Are Wrong: Option 6.5 hours would correspond to a closing distance of 10 * 6.5 = 65 km, which is more than the initial 60 km and therefore inconsistent. Option 6.2 hours and 6.52 hours likewise give closing distances that do not match the exact initial gap of 60 km. Only 6 hours gives a perfect match and satisfies the relationship between distance, time and relative speed. Common Pitfalls: Learners sometimes add the speeds instead of subtracting them when the vehicles are moving in the same direction, which is incorrect. Another common mistake is to mix up which car is ahead and which is behind, but the relative speed formula using the difference in speeds works regardless, as long as we use the absolute difference. It is also important to keep units consistent and to avoid guessing without using the distance divided by relative speed formula. Final Answer: The faster car will catch up with the slower car after 6 hours.

Question 2

A man covers a total distance of 100 km on a bicycle. For the first 2 hours he rides at 20 km/h and for the remaining journey he rides at 10 km/h. What is his average speed for the whole journey in km/h?

Accepted Answer

Introduction / Context: This problem tests your understanding of the concept of average speed when different parts of a journey are travelled at different speeds. Many students mistakenly take the simple average of the given speeds, but that is not correct unless the times or distances satisfy specific conditions. Here, the man rides at one speed for a fixed time and then switches to a different speed for the rest of the distance. We must compute the total time and then use the definition of average speed. Given Data / Assumptions: Total distance covered = 100 km. Speed for the first part of the trip = 20 km/h. Duration of the first part = 2 hours. Speed for the remaining distance = 10 km/h. Motion is uniform at each stated speed and there are no breaks. Concept / Approach: The average speed for the entire journey is defined as total distance divided by total time. We first compute how much distance is covered in the first 2 hours at 20 km/h. The remaining distance is then covered at 10 km/h. From this we calculate the time taken for the second part, add both times to obtain total time, and finally divide the total distance of 100 km by the total time. This correct approach avoids the mistake of simply averaging the two speeds. Step-by-Step Solution: Distance covered in the first 2 hours = 20 * 2 = 40 km. Remaining distance = total distance 100 km - 40 km = 60 km. Time taken to cover the remaining 60 km at 10 km/h = 60 / 10 = 6 hours. Total time for the whole journey = 2 hours + 6 hours = 8 hours. Average speed = total distance / total time = 100 / 8 km/h. Compute 100 / 8 = 12.5 km/h. Verification / Alternative check: We can double check by thinking about the relative contributions of the speeds. The man spends only 2 hours at the higher speed and as many as 6 hours at the lower speed. Therefore the average speed must be much closer to 10 km/h than to 20 km/h. A value like 12.5 km/h is reasonable. If we had simply averaged 20 and 10 to get 15 km/h, that would ignore the fact that he spent more time at the slower speed, which would not make sense. Why Other Options Are Wrong: Option 13 km/h is too high given that most of the time is at 10 km/h. Option 15.5 km/h and 20 km/h are even more unrealistic because they imply that the slow part has little influence, which is not true. Only 12.5 km/h matches the calculation based on total distance and total time, making it the correct average speed for the complete trip. Common Pitfalls: A very common mistake is to take the simple arithmetic mean of the two speeds without considering how long each speed was maintained. Another error is to forget that total distance is fixed at 100 km and assume equal distances for each speed, which is not given. Learners may also miscalculate the remaining distance or forget to convert all quantities into consistent units before performing the calculations. Final Answer: The average speed of the man for the entire bicycle journey is 12.5 km/h.

Question 3

Sound travels through air at approximately 330 metres per second. If the sound of thunder is heard 10 seconds after the flash of lightning, what is the approximate distance in kilometres of the thunder cloud?

Accepted Answer

Introduction / Context: This question connects a simple time, speed and distance relationship to a real life phenomenon. When we see lightning and hear thunder after a delay, the time difference allows us to estimate how far away the thunder cloud is, because light travels almost instantaneously compared to sound. The problem gives the speed of sound and the time delay and asks for the distance of the cloud in kilometres. Given Data / Assumptions: Speed of sound in air = 330 metres per second. Time delay between flash and thunder = 10 seconds. Speed is assumed constant throughout the travel of the sound. We neglect the time taken by light to reach us because it is extremely small compared to sound. We must report distance in kilometres. Concept / Approach: We use the basic formula distance = speed * time. First we compute the distance in metres by multiplying speed and time. Then we convert this distance into kilometres by dividing by 1000, because one kilometre equals 1000 metres. The result gives the approximate distance of the thunder cloud from the observer. This type of problem often appears in physics based aptitude tests and helps build comfort with unit conversions. Step-by-Step Solution: Distance in metres = speed * time = 330 m/s * 10 s. Compute 330 * 10 = 3300 metres. Convert metres into kilometres: distance in km = 3300 / 1000. 3300 / 1000 = 3.3 kilometres. Verification / Alternative check: As a quick mental check, if the delay were 3 seconds, the distance would be roughly 1 kilometre because 330 * 3 = 990 metres which is close to 1 km. For 10 seconds, we therefore expect a little more than 3 km. Our calculated value of 3.3 km fits this reasoning perfectly, reinforcing confidence in the numerical result. Why Other Options Are Wrong: Option 0.33 km would correspond to a distance of only 330 metres, which would require a delay of about 1 second, not 10 seconds. Option 33 km and 33.3 km are ten times larger than the correct distance and would require a delay of roughly 100 seconds at the given speed. They do not match the provided 10 second delay. Only 3.3 km is consistent with distance = 330 * 10 metres converted to kilometres. Common Pitfalls: Common mistakes include forgetting to convert metres into kilometres or misplacing the decimal point when dividing by 1000. Some learners also mistakenly multiply 330 by 100 instead of 10, or misread the speed as kilometres per hour. Always pay attention to the units given and perform conversions step by step to avoid errors. Final Answer: The thunder cloud is approximately 3.3 kilometres away from the observer.

Question 4

Gautam goes from his home to his office at 12 km/h and returns from office to home at 10 km/h by the same route. What is his average speed for the entire journey in km/h?

Accepted Answer

Introduction / Context: This problem is about average speed for a round trip when the speeds in each direction are different but the distance is the same. Many candidates mistakenly take the simple arithmetic mean of the two speeds, but the correct average speed in such cases is the harmonic mean because the time spent at each speed is different. This question helps test and reinforce that important concept. Given Data / Assumptions: Speed from home to office = 12 km/h. Speed from office to home = 10 km/h. The route and distance each way are the same. Let the one way distance be D kilometres. Motion is uniform at each stated speed. Concept / Approach: Average speed is defined as total distance divided by total time. For a journey where the distance each way is the same but speeds differ, a convenient formula is harmonic mean: average speed = 2 * v1 * v2 / (v1 + v2). This comes directly from computing total distance 2D and total time D / v1 + D / v2. After simplifying, D cancels out and we obtain the harmonic mean expression. We apply this formula with v1 = 12 km/h and v2 = 10 km/h. Step-by-Step Solution: Let the one way distance be D km (it will cancel later). Time from home to office = D / 12 hours. Time from office to home = D / 10 hours. Total distance = 2D km and total time = D/12 + D/10 hours. Total time = D(1/12 + 1/10) = D((5 + 6)/60) = 11D/60 hours. Average speed = total distance / total time = 2D / (11D/60) = 2D * 60 / (11D) = 120/11 km/h. Compute 120/11 ≈ 10.9 km/h. Verification / Alternative check: We can also compute using an assumed distance. Suppose D = 60 km. Time going = 60 / 12 = 5 hours. Time returning = 60 / 10 = 6 hours. Total distance = 120 km and total time = 11 hours, so average speed = 120 / 11 ≈ 10.9 km/h. The value matches the harmonic mean formula and confirms the result. Importantly, the answer is closer to the lower speed because more time is spent at the slower speed. Why Other Options Are Wrong: Option 11 km/h is close but slightly higher than the correct harmonic mean. Option 22 km/h is obviously too high, roughly double the individual speeds. Option 12.5 km/h is the simple average of 12 and 10, but that formula does not apply when distances are equal and speeds differ because the times are not equal. Therefore only 10.9 km/h is consistent with distance and time calculations. Common Pitfalls: The most common error is to compute (12 + 10)/2 and select 11 km/h. Another pitfall is forgetting that average speed depends on total distance and total time, not just the numerical values of speeds. Some learners also make algebraic mistakes while simplifying D / 12 + D / 10. Always factor D and carefully combine fractions before simplifying. Final Answer: Gautam’s average speed for the whole trip is approximately 10.9 km/h.

Question 5

A taxi moves at 40 km/h and takes 25 minutes to travel a certain distance. By how many kilometres per hour should it increase its speed in order to cover the same distance in 20 minutes?

Accepted Answer

Introduction / Context: This is another time, speed and distance question similar to earlier ones but with different numbers. The taxi initially travels at a given speed and takes a known time to cover a fixed distance. We are asked by how much the speed must be increased so that the same distance is covered in less time. This type of scenario helps you understand inverse relationships between speed and time when distance is constant. Given Data / Assumptions: Initial speed of the taxi = 40 km/h. Time taken at this speed = 25 minutes. Required time for the same distance = 20 minutes. The distance remains the same in both journeys. We must find the increase in speed in km/h. Concept / Approach: For constant distance, speed and time are inversely proportional. First we compute the distance covered in the initial case by converting time into hours and multiplying by the speed. Next we compute the speed required to cover this same distance in the shorter time. The difference between the new speed and the original speed gives the required increase. Careful unit conversion from minutes to hours is essential to avoid errors. Step-by-Step Solution: Convert 25 minutes into hours: 25 minutes = 25/60 hour. Original distance = 40 * (25/60) km = 40 * 25 / 60 = 1000 / 60 ≈ 16.666 km. Convert 20 minutes into hours: 20 minutes = 20/60 = 1/3 hour. Required new speed = same distance / new time = (1000/60) / (1/3) km/h. Simplify: (1000/60) * 3 = 1000 / 20 = 50 km/h. Increase in speed = 50 - 40 = 10 km/h. Verification / Alternative check: We can check by reasoning with ratios. Time is reduced from 25 minutes to 20 minutes. The ratio of times is 25:20 which simplifies to 5:4. For constant distance, the ratio of speeds must be the inverse, that is 4:5. Therefore the new speed must be 40 * 5/4 = 50 km/h. The difference between 50 and 40 is again 10 km/h, confirming our earlier calculation. Why Other Options Are Wrong: Option 50 represents the new speed, not the increase in speed, so it does not directly answer the question. Option 5 km/h and 25 km/h do not correspond to any correct ratio between old and new speed derived from the change in time. They would produce either too small or too large a new speed to achieve the required 20 minute travel time. Only an increase of 10 km/h results in the correct new speed of 50 km/h. Common Pitfalls: Students sometimes forget to convert minutes into hours and directly multiply 40 by 25, leading to an incorrect distance. Another mistake is to treat the percentage change in time as the same as the percentage change in speed without using the inverse relationship. It is also possible to confuse the new speed with the increase in speed. Always compute both clearly and then subtract to find the increase. Final Answer: The taxi must increase its speed by 10 km/h.

Question 6

A person walks at 12 km/h instead of 10 km/h and would then cover 1 kilometre more in the same time. What is the actual distance in kilometres that he walks at 10 km/h in that time?

Accepted Answer

Introduction / Context: This problem is another example of how changing speed while keeping time constant affects the distance travelled. It states that if the person walks faster, he covers 1 kilometre more in the same time. From this information we must find the actual distance he covers at the lower speed. Such questions are important for building algebraic thinking using the basic speed, time and distance formula. Given Data / Assumptions: Lower walking speed = 10 km/h. Higher walking speed = 12 km/h. The time of travel is the same in both cases. At the higher speed he covers 1 km more than at the lower speed. We must find the distance covered at 10 km/h. Concept / Approach: Using distance = speed * time, we express the distance at each speed in terms of a common time variable. The difference between these distances is given as 1 km. This yields a simple linear equation in the time variable. Once we find the time, we can compute the distance covered at 10 km/h. This algebraic method avoids guesswork and works systematically for all similar problems. Step-by-Step Solution: Let t be the time in hours for which he walks. Distance at 10 km/h = 10t kilometres. Distance at 12 km/h = 12t kilometres. Given that the faster walk covers 1 km more: 12t - 10t = 1. So 2t = 1, which gives t = 1/2 hour. Distance at 10 km/h = 10t = 10 * 1/2 = 5 kilometres. Verification / Alternative check: We can verify numerically. If t = 0.5 hours, distance at 10 km/h is 10 * 0.5 = 5 km. Distance at 12 km/h is 12 * 0.5 = 6 km. The difference is 6 - 5 = 1 km, which matches exactly the condition in the question. Therefore t and the distance at 10 km/h are correctly calculated, and 5 km is the required answer. Why Other Options Are Wrong: If the distance were 8 km at 10 km/h, the time would be 8 / 10 = 0.8 hours and at 12 km/h the distance would be 9.6 km, giving a difference of 1.6 km, not 1 km. Similarly, if the distance were 10 km, the difference between 12t and 10t would not be 1 km. A distance of 12 km would imply an even larger difference. Only a distance of 5 km leads to exactly a 1 km increase when the speed rises from 10 to 12 km/h. Common Pitfalls: One common mistake is to assume that 2 km/h extra speed directly adds 2 km to the distance, which is not true because time also matters. Others attempt to guess the distance instead of forming an equation, leading to confusion. Some learners also forget that the time is the same in both cases and try to set up two separate unrelated expressions. Final Answer: The person actually walks 5 kilometres at 10 km/h in that time.

Question 7

The distance between the points with coordinates (4, 8) and (k, -4) is 13 units. What is the possible value of k?

Accepted Answer

Introduction / Context: This question checks understanding of the distance formula in coordinate geometry and the ability to work with algebraic expressions. We are given two points in the coordinate plane, one of which has an unknown x coordinate k, and we are told that the distance between these points is 13 units. Using the distance formula and the given coordinates, we must find the correct value of k that satisfies this geometric condition. This type of problem commonly appears in school mathematics and aptitude tests to reinforce algebra and geometry together. Given Data / Assumptions: First point A has coordinates (4, 8). Second point B has coordinates (k, -4). The distance between A and B is 13 units. We assume a standard Cartesian coordinate system. We need the value of k that satisfies the distance formula. Concept / Approach: In a two dimensional Cartesian plane, the distance between points (x1, y1) and (x2, y2) is given by the formula distance^2 = (x2 - x1)^2 + (y2 - y1)^2. If the distance is known, we can square it and set up an equation: (x2 - x1)^2 + (y2 - y1)^2 = given distance squared. In this problem we substitute x1 = 4, y1 = 8, x2 = k and y2 = -4, then equate the expression to 13^2. Solving the resulting quadratic equation in k gives possible values for k. Finally, we choose the value that appears in the options. Step-by-Step Solution: Point A = (4, 8) and point B = (k, -4). Using the distance formula: (k - 4)^2 + (-4 - 8)^2 = 13^2. Compute the y difference: -4 - 8 = -12, so (-12)^2 = 144. So (k - 4)^2 + 144 = 169. Subtract 144 from both sides: (k - 4)^2 = 169 - 144 = 25. Therefore k - 4 = 5 or k - 4 = -5, which gives k = 9 or k = -1. Verification / Alternative check: We now check which of these k values appear in the options. The options include -1 but do not include 9. For k = -1, the points become (4, 8) and (-1, -4). The horizontal difference is 4 - (-1) = 5 units and the vertical difference is 8 - (-4) = 12 units. The distance squared is 5^2 + 12^2 = 25 + 144 = 169, so the distance is 13 units. This confirms that k = -1 satisfies the condition and is compatible with the given choices. Why Other Options Are Wrong: If k = 1, then (k - 4)^2 = (-3)^2 = 9, and the total distance squared would be 9 + 144 = 153, giving a distance less than 13. For k = 3, we would get (3 - 4)^2 = 1, and the distance squared would become 1 + 144 = 145, still not 169. For k = -3, we get (-3 - 4)^2 = (-7)^2 = 49, and the distance squared would be 49 + 144 = 193, which is greater than 169. Therefore none of these values produce 13 as the distance. Common Pitfalls: Common mistakes include forgetting to square the differences correctly, especially the vertical difference, or forgetting that we must square the distance 13 to obtain 169 in the equation. Some learners also mistake 13^2 for 13 * 2 or incorrectly handle negative signs when computing differences like -4 - 8. Another pitfall is not checking which roots of the quadratic equation are actually present in the options and accidentally using a value not listed. Final Answer: The valid value of k from the given options is -1.

Question 8

A spaceship travels in space at a constant speed of 1 260 km/h. How many metres does it travel in one tenth of a second?

Accepted Answer

Introduction / Context: Here we apply speed conversion and the distance formula in a context involving a spaceship. The question gives a high speed in kilometres per hour and asks for the distance covered in a very small time interval, one tenth of a second. This tests the ability to convert between kilometres per hour and metres per second and then apply distance = speed * time correctly. Given Data / Assumptions: Speed of the spaceship = 1 260 km/h. Time interval considered = 1/10 second. One kilometre = 1 000 metres. One hour = 3 600 seconds. Motion is uniform at the given speed. Concept / Approach: We first convert the speed from km/h into m/s, using the standard conversion factor. Once we know the speed in metres per second, the distance travelled in any time interval in seconds is computed as distance = speed * time. Since the time is 1/10 second, we simply multiply the speed in m/s by 0.1. Finally, we compare the result with the given options to select the correct distance. Step-by-Step Solution: Speed = 1 260 km/h. Convert to m/s using 1 km/h = 5/18 m/s. Speed in m/s = 1 260 * 5/18. Compute: 1 260 / 18 = 70, so speed = 70 * 5 = 350 m/s. Distance travelled in 1/10 second = 350 * (1/10) = 35 metres. Verification / Alternative check: We can also compute directly using definitions. In one hour the spaceship travels 1 260 km, which is 1 260 000 metres. In 3 600 seconds it therefore covers 1 260 000 metres. Distance per second is 1 260 000 / 3 600 = 350 metres. Then in 0.1 second the distance is 350 * 0.1 = 35 metres. This exactly matches our earlier calculation and confirms the result as 35 metres. Why Other Options Are Wrong: Option 126 metres would correspond to distance covered in roughly 0.36 seconds at 350 m/s. Option 36 metres is very close but slightly higher than 35 and would correspond to a slightly longer time than 1/10 second. Option 125 metres is far too large for such a short time at the given speed. Only 35 metres accurately reflects the distance travelled in one tenth of a second. Common Pitfalls: A common error is to multiply by 18/5 instead of 5/18 when converting from km/h to m/s, which produces a very large and incorrect speed. Some learners also forget that 1/10 second equals 0.1 second and accidentally treat it as 10 seconds. Misplacing zeroes when converting kilometres into metres can lead to errors as well. Keeping track of units at every step helps prevent these mistakes. Final Answer: The spaceship travels 35 metres in one tenth of a second.

Question 9

A spaceship travels in space at a constant speed of 810 km/h. How many metres does it travel in one fifth of a second?

Accepted Answer

Introduction / Context: This is another speed conversion problem involving a spaceship moving at high speed. We are given the speed in kilometres per hour and asked for the distance travelled in one fifth of a second. This tests familiarity with converting km/h to m/s and then applying the distance formula for a small time interval. Given Data / Assumptions: Speed of the spaceship = 810 km/h. Time interval = 1/5 second. 1 kilometre = 1 000 metres. 1 hour = 3 600 seconds. Motion is uniform at the stated speed. Concept / Approach: We again use the standard conversion factor between kilometres per hour and metres per second. Once the speed is known in metres per second, the distance travelled in any number of seconds is simply speed multiplied by time. Since the time is 1/5 second, we multiply the computed m/s value by 0.2. This method is quick and reliable for all similar questions. Step-by-Step Solution: Speed = 810 km/h. Convert to m/s using 1 km/h = 5/18 m/s. Speed in m/s = 810 * 5/18. Compute 810 / 18 = 45, so speed = 45 * 5 = 225 m/s. Time interval = 1/5 second = 0.2 second. Distance = speed * time = 225 * 0.2 = 45 metres. Verification / Alternative check: Using the definition, in one hour the spaceship travels 810 km = 810 000 metres. In 3 600 seconds this is 810 000 metres, so per second it travels 810 000 / 3 600 = 225 metres. In 0.2 seconds the distance is 225 * 0.2 = 45 metres. This confirms the earlier calculation and shows that 45 metres is correct. Why Other Options Are Wrong: Option 60 metres would be the distance travelled in about 0.266 seconds at 225 m/s, not 0.2 seconds. Option 75 metres corresponds to 0.333 seconds at 225 m/s. Option 90 metres corresponds to 0.4 seconds. None of these match the exact 1/5 second interval. Only 45 metres exactly fits the relationship between speed and time given in the question. Common Pitfalls: Learners may misread 1/5 as 5 seconds or confuse it with 0.05. Another common error is to use 18/5 instead of 5/18 for the conversion, which yields unrealistic distances. It is also easy to mistype or miscalculate 810 * 5/18 without reducing the fraction first. Simplifying 810 / 18 to 45 before multiplying makes the arithmetic easier and clearer. Final Answer: The spaceship travels 45 metres in one fifth of a second.

Question 10

A car has to cover a distance of 144 km in 3.2 hours. What average speed in metres per second should the car maintain to cover this distance on time?

Accepted Answer

Introduction / Context: This question combines the idea of average speed with unit conversion from kilometres per hour to metres per second. The car must cover a fixed distance in a fixed time, and we are asked to express the required average speed in m/s rather than km/h. Such problems are very common in exams and also reflect how speed is measured in different contexts such as road travel and physics. Given Data / Assumptions: Required distance = 144 km. Available time = 3.2 hours. We need average speed in metres per second. 1 km = 1 000 m and 1 hour = 3 600 s. Car travels at a constant average speed over the entire distance. Concept / Approach: The average speed in km/h is distance divided by time, both taken in kilometres and hours. Once that value is obtained, we convert it into metres per second using the standard factor 5/18. It is important to treat 3.2 hours as a decimal number of hours, equal to 3 hours and 12 minutes, but we can work directly in decimal form without converting to minutes. Step-by-Step Solution: Average speed in km/h = distance / time = 144 / 3.2. Compute 144 / 3.2. Multiply numerator and denominator by 10: 1440 / 32 = 45 km/h. Convert 45 km/h to m/s using factor 5/18. Speed in m/s = 45 * 5/18. Compute: 45 / 9 = 5, so 45 * 5/18 = (45/18) * 5 = 2.5 * 5 = 12.5 m/s. Verification / Alternative check: We can verify by converting 45 km/h directly. In one hour at 45 km/h, the car travels 45 000 metres. Dividing by 3 600 seconds in an hour gives 45 000 / 3 600 = 12.5 metres per second. This matches the value obtained using the 5/18 factor. Therefore 12.5 m/s is the correct average speed required to cover 144 km in 3.2 hours. Why Other Options Are Wrong: Option 10 m/s corresponds to only 36 km/h, which would be too slow to cover 144 km in 3.2 hours. Option 7.5 m/s corresponds to 27 km/h, even slower. Option 15 m/s corresponds to 54 km/h, which would cover the distance in less than 3.2 hours. Only 12.5 m/s corresponds to exactly 45 km/h, the speed that satisfies the given distance and time requirement. Common Pitfalls: Common mistakes include misreading 3.2 hours as 3 hours 20 minutes rather than 3 hours 12 minutes, or incorrectly computing 144 / 3.2. Some students try to convert units first and then divide, which can be more complicated. It is simpler to find the speed in km/h and then convert. Errors also occur when using the conversion factor, such as multiplying by 18/5 instead of 5/18. Final Answer: The car should maintain an average speed of 12.5 metres per second.

Question 11

A car travels from town A to town B at 40 km/h and returns from B to A at 60 km/h by the same route. What is the average speed of the car for the entire journey?

Accepted Answer

Introduction / Context: This is a standard round trip average speed problem where the distances each way are equal but the speeds are different. Many candidates incorrectly choose the simple average of 40 and 60, but the correct average speed must take into account the total time taken for the journey. This question reinforces the formula for average speed in such two way journeys, which is based on the harmonic mean of the speeds. Given Data / Assumptions: Speed from A to B = 40 km/h. Speed from B to A = 60 km/h. The distance between A and B is the same in both directions. Let the one way distance be D kilometres. We assume uniform motion at the stated speeds. Concept / Approach: Average speed equals total distance divided by total time. If the distances are equal, we can use the harmonic mean formula for speeds v1 and v2: average speed = 2 * v1 * v2 / (v1 + v2). This comes from writing the total time as D / v1 + D / v2 and total distance as 2D. Simplifying 2D divided by that total time yields the harmonic mean. We apply this formula with v1 = 40 and v2 = 60. Step-by-Step Solution: Let the one way distance be D km. Time from A to B = D / 40 hours. Time from B to A = D / 60 hours. Total distance = 2D km. Total time = D/40 + D/60 = D(1/40 + 1/60) = D((3 + 2)/120) = 5D/120 = D/24 hours. Average speed = total distance / total time = 2D / (D/24) = 2D * 24 / D = 48 km/h. Verification / Alternative check: Assume D = 120 km for easier numbers. Time going at 40 km/h is 120 / 40 = 3 hours. Time returning at 60 km/h is 120 / 60 = 2 hours. Total distance = 240 km and total time = 5 hours. Average speed = 240 / 5 = 48 km/h, which matches our formula based solution. This confirms that 48 km/h is correct. Why Other Options Are Wrong: Option 50 km/h is simply a guess and does not arise from any correct calculation. Option 45 km/h is the wrong type of average and is lower than the true value. Option 60 km/h is just one of the given speeds and cannot be the average when part of the journey is at a slower 40 km/h. Only 48 km/h correctly balances the distances and times for the full round trip. Common Pitfalls: The main pitfall is using the arithmetic mean (40 + 60)/2 = 50 km/h as the answer. That approach ignores the different times spent at each speed. Another mistake is to miscalculate the combined time when working with fractions, for example adding 1/40 and 1/60 incorrectly. Using an assumed numerical distance such as 120 km often makes the arithmetic clearer and reduces algebraic errors. Final Answer: The average speed of the car for the entire journey is 48 km/h.

Question 12

An aeroplane has to cover a distance of 1 200 km. Due to bad weather it reduces its speed by 300 km/h and therefore reaches the destination 2 hours late. What was the scheduled duration of the flight in hours under normal speed conditions?

Accepted Answer

Introduction / Context: This question involves forming an equation using the time, speed and distance relationship when the speed of an aeroplane changes. The plane was supposed to fly at a higher normal speed but due to bad weather it reduces its speed by a fixed amount and arrives 2 hours later than scheduled. We are asked to find the originally planned duration of the flight, not the speed directly. This is a classic algebraic application in time and distance. Given Data / Assumptions: Planned distance = 1 200 km. Normal speed = v km/h (unknown). Reduced speed = v - 300 km/h. Distance remains 1 200 km in both scenarios. Actual flight time = scheduled time + 2 hours. Concept / Approach: Time taken for a given distance is distance / speed. Let the scheduled time be T hours, so T = 1 200 / v. With reduced speed the time becomes 1 200 / (v - 300). We are told that the new time is 2 hours more than the original time. Therefore, 1 200 / (v - 300) = 1 200 / v + 2. This equation can be solved for v and then we can compute T. Since the question asks for the duration, our final answer is T in hours. Step-by-Step Solution: Let normal speed be v km/h. Scheduled time T = 1 200 / v hours. Reduced speed = v - 300 km/h, so delayed time = 1 200 / (v - 300) hours. Given: delayed time = scheduled time + 2 hours. So 1 200 / (v - 300) = 1 200 / v + 2. Solve this equation: rearrange and solve to get v = 600 km/h. Then scheduled time T = 1 200 / 600 = 2 hours. Verification / Alternative check: With v = 600 km/h, the normal time is 1 200 / 600 = 2 hours. The reduced speed is 600 - 300 = 300 km/h. At this reduced speed, time = 1 200 / 300 = 4 hours. The difference between the delayed time and the scheduled time is 4 - 2 = 2 hours, which matches the given condition. Therefore the scheduled duration of 2 hours is correct. Why Other Options Are Wrong: If the scheduled time were 1 hour, the normal speed would need to be 1 200 km/h, giving a reduced speed of 900 km/h which would not result in only 2 hours extra. A scheduled time of 1.5 hours or 2.5 hours leads to speeds and delayed times that do not give a difference of exactly 2 hours. Only a scheduled duration of 2 hours corresponds to a realistic speed of 600 km/h and a reduced speed of 300 km/h that fits all the conditions. Common Pitfalls: Some learners directly try to guess the speed or time without forming an equation, which can lead to confusion. Others forget to subtract 300 from the normal speed and instead divide by 300, which is incorrect. It is also easy to make algebraic mistakes while solving 1 200 / (v - 300) = 1 200 / v + 2. Writing each step clearly and checking the resulting speed and times helps avoid errors. Final Answer: The scheduled duration of the flight under normal conditions was 2 hours.

Question 13

A square playground has an area of 1 127.6164 square metres. If a man walks at a speed of 2 9/20 metres per minute, approximately how many minutes will he take to walk one complete round around the boundary of the playground?

Accepted Answer

Introduction / Context: This question combines geometry and time and distance concepts. We are given the area of a square playground and the walking speed of a man in metres per minute. The task is to find the approximate time he takes to walk one complete lap around the boundary of the square. That means we must move from area to side length, then to perimeter, and finally relate perimeter to speed and time. Such multi step problems are very common in aptitude tests because they check multiple concepts together. Given Data / Assumptions: The playground is exactly square in shape. Area of the square = 1 127.6164 square metres. Walking speed of the man = 2 9/20 metres per minute. 2 9/20 metres per minute can be converted to an improper fraction. We need the approximate time in minutes to walk once around the playground. Concept / Approach: For a square, area = side^2, so the side length is the square root of the area. Once we know the side length, we can compute the perimeter as 4 times the side. The total distance walked in one lap is this perimeter. Using the time, speed and distance relationship distance = speed * time, we rearrange to time = distance / speed. Because the numbers are not very simple, we approximate the square root and the final time to two decimal places, matching the style of the given options. Step-by-Step Solution: Given area of the square = 1 127.6164 square metres. Side length = square root of 1 127.6164 ≈ 33.58 metres. Perimeter of the square = 4 * side ≈ 4 * 33.58 = 134.32 metres. Walking speed = 2 9/20 metres per minute. Convert 2 9/20 to an improper fraction: 2 9/20 = (2 * 20 + 9) / 20 = 49/20 = 2.45 metres per minute. Time taken to walk one round = perimeter / speed ≈ 134.32 / 2.45 minutes. Compute 134.32 / 2.45 ≈ 54.82 minutes (approximately). Verification / Alternative check: To check the approximation, we can multiply the speed back by the time. Using 54.82 minutes at 2.45 metres per minute gives a distance of approximately 54.82 * 2.45 ≈ 134.32 metres, which matches the calculated perimeter. Also, 33.58 squared is very close to 1 127.6164, confirming that our side length is accurate. These checks show that the approximations are consistent and that 54.82 minutes is a reliable value. Why Other Options Are Wrong: Option 50.82 minutes corresponds to a shorter total distance when multiplied by 2.45 and would not reach the full perimeter of about 134.32 metres. Option 54.62 minutes is slightly smaller than 54.82 and leads to a smaller distance than needed. Option 50.62 minutes is also too small. Only 54.82 minutes, when multiplied by the walking speed, correctly reproduces the perimeter of the playground and aligns with the geometry of the square. Common Pitfalls: Learners often forget that the given area is for a square and attempt to use formulas for rectangles or circles. Another common error is to miscalculate the square root or to approximate it too coarsely, which leads to incorrect perimeters. Mistakes also occur when converting the mixed fraction 2 9/20 into an improper fraction or decimal. Carefully carrying each step, maintaining sufficient decimal accuracy and checking by multiplying back are good habits to avoid these issues. Final Answer: The man will take approximately 54.82 minutes to walk one round around the playground.

Question 14

Two donkeys are standing 400 metres apart. The first donkey can run at 3 m/s and the second can run at 2 m/s. If both donkeys start running towards each other at the same time, after how many seconds will they bump into each other?

Accepted Answer

Introduction / Context: This problem is a simple example of relative speed when two objects move towards each other. The donkeys start some distance apart and run directly towards one another at given speeds. We are asked to find how long it takes before they meet. Such questions are very common in aptitude tests and train quick reasoning with relative motion. Given Data / Assumptions: Initial distance between the two donkeys = 400 metres. Speed of the first donkey = 3 m/s. Speed of the second donkey = 2 m/s. They start at the same time and run directly towards each other. They run in a straight line at constant speeds. Concept / Approach: When two objects move towards each other, their relative speed is the sum of their individual speeds. The distance between them decreases at this combined speed. Time taken to meet is therefore the initial distance divided by the relative speed. We simply add the two speeds, then divide the initial distance by this sum to get the meeting time. Step-by-Step Solution: Initial distance = 400 metres. Speed of first donkey = 3 m/s. Speed of second donkey = 2 m/s. Relative speed towards each other = 3 + 2 = 5 m/s. Time to meet = distance / relative speed = 400 / 5 seconds. Compute 400 / 5 = 80 seconds. Verification / Alternative check: We can check by computing how far each donkey runs. In 80 seconds, the first donkey covers 3 * 80 = 240 metres and the second covers 2 * 80 = 160 metres. The sum 240 + 160 = 400 metres, exactly the initial separation. This confirms that 80 seconds is the correct time for them to bump into each other. Why Other Options Are Wrong: If the time were 60 seconds, the total distance covered would be (3 + 2) * 60 = 300 metres, which is not enough to close the 400 metre gap. If the time were 40 seconds, only 200 metres would be covered in total. Option 400 seconds is far too long and would correspond to a total distance of 2 000 metres at the relative speed of 5 m/s. Only 80 seconds gives exactly 400 metres, matching the initial separation. Common Pitfalls: Students sometimes subtract the speeds instead of adding them when the objects are moving towards each other, which is incorrect. Another error is to confuse metres per second with kilometres per hour and attempt unnecessary conversions. Remember that for motion towards each other, relative speed is the sum of speeds when they are along the same line. Final Answer: The donkeys will bump into each other after 80 seconds.

Question 15

Rubi walks from her home to a multiplex. At a speed of 3 km/h she reaches 5 minutes late, while at a speed of 4 km/h she reaches 5 minutes early. What is the distance in kilometres between her home and the multiplex?

Accepted Answer

Introduction / Context: This question is very similar to the earlier example with Ram and his school. It involves early and late arrival times for two different walking speeds. Such problems test your ability to convert minutes into hours, form equations using time = distance / speed, and use the difference in times to deduce the unknown distance. These patterns repeat often in competitive examinations. Given Data / Assumptions: Speed in the first case = 3 km/h and Rubi is 5 minutes late. Speed in the second case = 4 km/h and Rubi is 5 minutes early. Let the scheduled travel time be T hours. Let the distance between home and the multiplex be D kilometres. The path is the same and speeds are constant in each case. Concept / Approach: Again, we use time = distance / speed. At the lower speed, the time taken equals the scheduled time plus the amount of lateness. At the higher speed, the time taken equals the scheduled time minus the amount by which she is early. The difference between these two times equals the total difference between late and early arrivals. Solving the resulting equation allows us to find D directly. Step-by-Step Solution: Let D be the distance in km and T be the scheduled travel time in hours. At 3 km/h, actual time = D / 3 = T + 5/60. At 4 km/h, actual time = D / 4 = T - 5/60. Subtract second equation from first: D/3 - D/4 = (T + 5/60) - (T - 5/60). Left side: D/3 - D/4 = (4D - 3D) / 12 = D / 12. Right side: T cancels, giving 5/60 + 5/60 = 10/60 = 1/6 hour. So D / 12 = 1/6, therefore D = 12 * 1/6 = 2 kilometres. Verification / Alternative check: We can verify by computing the actual times. If D = 2 km, at 3 km/h time = 2 / 3 hour = 40 minutes. At 4 km/h time = 2 / 4 hour = 30 minutes. The difference between these times is 10 minutes. That exactly matches the total variation, since Rubi is 5 minutes late in one case and 5 minutes early in the other, giving a total difference of 10 minutes. Thus the value D = 2 km fits the entire situation correctly. Why Other Options Are Wrong: If the distance were 5 km, the times at 3 and 4 km/h would differ by far more than 10 minutes. Distances of 2.5 km or 5.5 km would similarly produce mismatched time differences. Only 2 km gives precisely a 10 minute difference between the two travel times and matches the early and late conditions described in the question. Common Pitfalls: Learners sometimes treat the difference between late and early as 0 minutes instead of adding the two 5 minute intervals. Another mistake is to convert minutes incorrectly or to forget that 5 minutes equals 5/60 hours. It is also easy to attempt guessing the distance rather than setting up an equation, which becomes unreliable as the numbers change. Systematic use of time = distance / speed is the safest method. Final Answer: The distance between Rubi’s home and the multiplex is 2 kilometres.

Question 16

A man travels from one place to another at 12 km/h and returns along the same route at 9 km/h. If the total time taken for the two way journey is 2 hours 20 minutes, what is the distance between the two places in kilometres?

Accepted Answer

Introduction / Context: This problem is another round trip time and distance question, but here the total time is given and the distance is unknown. The man travels one way at 12 km/h and returns at 9 km/h. We must use the total time of 2 hours 20 minutes to find the one way distance between the two places. This requires forming and solving an equation using time = distance / speed. Given Data / Assumptions: Speed going = 12 km/h. Speed returning = 9 km/h. Total time for going and returning = 2 hours 20 minutes. 2 hours 20 minutes = 2 + 20/60 = 7/3 hours. The distance between the two places is the same each way, call it D kilometres. Concept / Approach: Time taken to go one way is distance divided by speed. Therefore, time while going is D / 12 and time while returning is D / 9. The sum of these two times equals the total time 7/3 hours. We then solve the equation D / 12 + D / 9 = 7/3 to find D. This is a standard algebraic manipulation of fractions and is a very common pattern in speed and time questions. Step-by-Step Solution: Let D be the one way distance in kilometres. Time going = D / 12 hours. Time returning = D / 9 hours. Total time = D/12 + D/9 = 7/3 hours. Find a common denominator: 12 and 9 have LCM 36. So D/12 = 3D/36 and D/9 = 4D/36, giving (3D + 4D)/36 = 7D/36. Thus 7D/36 = 7/3. Multiply both sides by 36: 7D = 7 * 12 = 84. Divide both sides by 7: D = 84 / 7 = 12 kilometres. Verification / Alternative check: Now check the times. At 12 km/h, time to cover 12 km is 12 / 12 = 1 hour. At 9 km/h, time to cover 12 km is 12 / 9 = 4/3 hours, which is 1 hour 20 minutes. Total time is 1 + 4/3 = 7/3 hours, equal to 2 hours 20 minutes, exactly matching the problem statement. This confirms that D = 12 km is correct. Why Other Options Are Wrong: If D were 21 km, the time going would be 21 / 12 and time returning 21 / 9, which add up to more than 3 hours. A distance of 9 km gives total time less than 2 hours 20 minutes. A distance of 35 km would produce an even larger total time. Only 12 km results in exactly 2 hours 20 minutes when using the given speeds of 12 km/h and 9 km/h. Common Pitfalls: Learners may incorrectly convert 2 hours 20 minutes into decimal hours, sometimes writing 2.2 instead of 2 + 20/60. Others add the speeds directly or take a simple average, which does not apply here because the distances are the same but speeds and times differ. Mismanaging fractions while adding D / 12 and D / 9 is also common. Keeping calculations clear and step by step prevents these errors. Final Answer: The distance between the two places is 12 kilometres.

Question 17

In a height and distance problem, the distance between two vertical pillars is 120 metres. The height of one pillar is three times the height of the other pillar. From the midpoint of the line segment joining their feet, the angles of elevation of t…

Accepted Answer

Introduction / Context: This question belongs to the standard height and distance category in quantitative aptitude. It combines geometry with trigonometry and uses the idea of complementary angles of elevation to relate the heights of two vertical pillars to the horizontal distance between them. Understanding how tangent behaves for complementary angles is the key idea tested here. Given Data / Assumptions: Distance between the feet of the two pillars = 120 metres. Heights of the pillars are in the ratio 1 : 3, so let the shorter pillar have height h metres and the taller pillar have height 3h metres. Observation point is the midpoint of the line joining the feet of the pillars, so it is 60 metres away from each pillar. Angles of elevation of the tops of the two pillars from the midpoint are complementary, so their sum is 90 degrees. Take √3 ≈ 1.732 where needed for numerical evaluation. Concept / Approach: The main concepts used are tangent of an angle in a right triangle and the property of complementary angles. At the midpoint, vertical height and horizontal distance form right triangles with the line of sight as the hypotenuse. For an angle θ, tan θ = opposite side / adjacent side. If two angles are complementary, say θ and 90 − θ, then tan θ × tan(90 − θ) = 1, because tan(90 − θ) = cot θ and tan θ × cot θ = 1. We use this property to form an equation in h, the unknown height of the shorter pillar, and then calculate the taller height as 3h. Step-by-Step Solution: Step 1: Let the shorter pillar have height h metres and the taller pillar have height 3h metres. The midpoint is 60 metres from each pillar. Step 2: Let the angle of elevation to the top of the shorter pillar be β. Then tan β = h / 60. Step 3: Let the angle of elevation to the top of the taller pillar be α. Then tan α = 3h / 60 = h / 20. Step 4: Given that the two angles are complementary, α + β = 90 degrees. Therefore, tan α × tan β = 1. Step 5: Substitute the values: (h / 20) * (h / 60) = 1, so 3h^2 / 3600 = 1, which simplifies to h^2 = 1200. Step 6: Hence h = √1200 = √(400 * 3) = 20√3 ≈ 20 * 1.732 = 34.64 metres. Step 7: The taller pillar has height 3h = 3 * 34.64 = 103.92 metres. Verification / Alternative check: We can quickly check the relation tan α × tan β = 1. Using h = 20√3, tan β = h / 60 = 20√3 / 60 = √3 / 3. Tan α = 3h / 60 = 60√3 / 60 = √3. Their product is (√3 / 3) * √3 = 3 / 3 = 1, which confirms that the corresponding angles are complementary as given in the question. The calculated height therefore satisfies both the ratio of heights and the trigonometric condition. Why Other Options Are Wrong: 34.64 metres gives the height of the shorter pillar, not the taller pillar required in the question. 51.96 metres and 69.28 metres correspond to 1.5h and 2h respectively and do not satisfy the given trigonometric condition when used as the taller height. The only value consistent with both the ratio and the complementary angle condition is 103.92 metres. Common Pitfalls: A common mistake is to forget that the midpoint is equidistant from both pillars and to use 120 metres instead of 60 metres as the horizontal distance in the tangent expressions. Another mistake is to assume that complementary angles directly force tan α = 1 / tan β without explicitly forming the product tan α × tan β = 1. Finally, some learners mistakenly use the shorter height instead of the taller height in the final answer. Careful reading of the question and correct formation of the trigonometric equation avoids these errors. Final Answer: The height of the taller pillar is 103.92 metres.

Question 18

Two vertical buildings have heights of 38 metres and 58 metres respectively. The straight-line distance between their tops is measured to be 52 metres. Using this information and applying the Pythagoras theorem, what is the horizontal distance (in m…

Accepted Answer

Introduction / Context: This question tests the learner on the concept of right triangles and the Pythagoras theorem, which is a standard topic in geometry and also appears frequently in quantitative aptitude. The heights of two buildings and the straight-line distance between their tops together form a right triangle with the horizontal ground distance between the buildings as the base and the difference in their heights as the vertical side. Given Data / Assumptions: Height of the first building = 38 metres. Height of the second building = 58 metres. Straight-line distance between the tops of the buildings = 52 metres. The buildings are vertical and stand on the same horizontal level ground. We assume the line joining the tops of the buildings and the line joining their bases form a right triangle with the vertical difference in height. Concept / Approach: If two vertical structures stand on a flat horizontal surface, the line connecting their tops and the line connecting their bases form the hypotenuse and the base of a right triangle. The difference in heights forms the vertical side. The Pythagoras theorem states that in a right triangle, hypotenuse^2 = base^2 + height^2. Here, the hypotenuse is the distance between the tops, the vertical side is the difference in heights, and the horizontal distance between the buildings is the unknown base we need to determine. Step-by-Step Solution: Step 1: Compute the difference in the heights of the two buildings: 58 − 38 = 20 metres. Step 2: Consider the right triangle where the hypotenuse is 52 metres and one side (vertical) is 20 metres. Step 3: Let the horizontal distance between the buildings be d metres. Then, by Pythagoras theorem, d^2 + 20^2 = 52^2. Step 4: Calculate d^2: 52^2 = 2704 and 20^2 = 400, so d^2 = 2704 − 400 = 2304. Step 5: Take the square root of 2304: d = √2304 = 48 metres. Verification / Alternative check: We can verify by checking Pythagoras theorem again. If the base is 48 metres and the vertical side is 20 metres, then hypotenuse^2 = 48^2 + 20^2 = 2304 + 400 = 2704. The square root of 2704 is 52, which matches the given distance between the tops of the buildings. This confirms that our computed horizontal distance of 48 metres is correct. Why Other Options Are Wrong: A distance of 42 metres or 44 metres would result in hypotenuse values that are not equal to 52 metres when substituted into the Pythagoras relation. A distance of 46 metres also fails to satisfy the equation d^2 + 20^2 = 52^2. Only d = 48 metres fits the numerical requirement exactly and is consistent with the geometry of the problem. Common Pitfalls: A frequent mistake is to add the heights instead of taking their difference, which would incorrectly use 38 + 58 instead of 58 − 38. Another error is to assume that the 52 metres is the base instead of the hypotenuse, which reverses the roles of the sides. Learners also sometimes miscalculate squares of numbers or square roots. Careful use of the Pythagoras theorem and accurate arithmetic leads to the correct result. Final Answer: The horizontal distance between the two buildings is 48 metres.

Question 19

A plane travels a distance of 1800 kilometres in 5 hours along a straight route. Using the basic speed formula and converting units appropriately, what is the average speed of the plane expressed in metres per second?

Accepted Answer

Introduction / Context: This problem checks understanding of the basic relationship between distance, time, and speed and the ability to convert speed from kilometres per hour to metres per second. Such questions are very common in quantitative aptitude tests and help reinforce the conversion factor between these two units of speed. Given Data / Assumptions: Total distance travelled by the plane = 1800 kilometres. Total time taken for the journey = 5 hours. The route is straight and the speed is assumed to be constant over the whole journey. We are required to express the final speed in metres per second, not in kilometres per hour. Use the standard conversion 1 kilometre = 1000 metres and 1 hour = 3600 seconds. Concept / Approach: The basic formula is speed = distance / time. First, we find the average speed in kilometres per hour by dividing the total distance by total time. Then we convert this value into metres per second. To convert from km/h to m/s, multiply the km/h value by 1000 and divide by 3600, which simplifies to dividing by 3.6. This two step approach gives a precise and reliable answer. Step-by-Step Solution: Step 1: Compute the speed in kilometres per hour: speed_kmph = distance / time = 1800 / 5 = 360 km/h. Step 2: Use the conversion relationship 1 km/h = 1000 metres / 3600 seconds = 1 / 3.6 m/s. Step 3: Convert 360 km/h to m/s: speed_mps = 360 / 3.6. Step 4: Evaluate 360 / 3.6 = 100, so the average speed is 100 metres per second. Step 5: Therefore, the plane moves at an average speed of 100 m/s over the entire journey. Verification / Alternative check: As a quick check, convert the total distance directly into metres and divide by the total time in seconds. Distance = 1800 km = 1800 * 1000 = 1800000 metres. Time = 5 hours = 5 * 3600 = 18000 seconds. Then speed = 1800000 / 18000 = 100 metres per second. This direct method confirms the result obtained using the conversion of km/h to m/s. Why Other Options Are Wrong: A value of 200 m/s would correspond to a much higher distance for the same time or a much shorter duration, which is not given. Values of 10 m/s or 20 m/s are too small; they would imply the plane travels only 180 or 360 kilometres in 5 hours. The value 50 m/s is exactly half of the correct speed and comes from incorrectly dividing by 7.2 instead of 3.6. Only 100 m/s matches both unit conversion and the given journey data. Common Pitfalls: Common mistakes include forgetting to convert hours into seconds, using 60 instead of 3600, or inverting the conversion factor. Some learners also accidentally multiply by 3.6 instead of dividing by 3.6 when converting from km/h to m/s. Keeping the relation straight, that km/h is a larger unit of speed than m/s for the same magnitude, helps to remember that numerical values decrease when converting from km/h to m/s. Final Answer: The average speed of the plane is 100 metres per second.

Question 20

Two bikers A and B start at the same time from two different points and ride towards each other along a straight road. Biker A rides at a constant speed of 75 km/h and biker B rides at a constant speed of 60 km/h. They meet exactly after 20 minutes.…

Accepted Answer

Introduction / Context: This question tests the concept of relative speed when two objects are moving towards each other. It is a standard type of time and distance problem that appears frequently in aptitude exams. The key skill is converting minutes to hours and correctly adding the individual speeds to obtain effective relative speed for objects moving in opposite directions. Given Data / Assumptions: Speed of biker A = 75 km/h. Speed of biker B = 60 km/h. They ride towards each other along a straight line. They meet after 20 minutes from the time they start. We assume their speeds remain constant during the entire journey. Concept / Approach: When two bodies move towards each other along a straight line, their relative speed is the sum of their individual speeds. Total distance between them is equal to relative speed multiplied by the time taken to meet. Since the speeds are given in km/h, the time must be converted into hours. The resulting product gives the initial separation distance between the two bikers. Step-by-Step Solution: Step 1: Convert the meeting time into hours: 20 minutes = 20 / 60 hours = 1 / 3 hour. Step 2: Compute the relative speed when they move towards each other: relative speed = 75 + 60 = 135 km/h. Step 3: Use the relation distance = speed * time. Here, distance between them initially = relative speed * time taken to meet. Step 4: Distance = 135 * (1 / 3) = 45 kilometres. Step 5: Therefore, the initial distance between biker A and biker B was 45 km. Verification / Alternative check: We can check by computing how far each biker travels in 1 / 3 hour. Biker A covers 75 * (1 / 3) = 25 km, and biker B covers 60 * (1 / 3) = 20 km. The sum 25 + 20 = 45 km matches the initial distance between them, confirming that our calculation is correct. The meeting point is somewhere between the two starting points such that these individual distances add up to the total separation. Why Other Options Are Wrong: If the distance were 60 km, at a combined speed of 135 km/h, they would need 60 / 135 hours, which is less than 20 minutes. A distance of 30 km or 15 km would mean even smaller meeting times than 20 minutes. A value of 40 km still does not match the given time when checked with the relative speed. Only 45 km gives exactly 20 minutes for them to meet. Common Pitfalls: Typical mistakes include subtracting speeds instead of adding them when the objects move towards each other, or forgetting to convert minutes into hours, which leads to large numerical errors. Some learners also apply distance = speed / time incorrectly instead of speed * time. Carefully identifying the direction of motion and ensuring consistent units avoids these errors. Final Answer: The bikers were initially separated by 45 kilometres.

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